/*
 * @lc app=leetcode.cn id=130 lang=cpp
 *
 * [130] 被围绕的区域
 *
 * https://leetcode.cn/problems/surrounded-regions/description/
 *
 * algorithms
 * Medium (45.67%)
 * Likes:    815
 * Dislikes: 0
 * Total Accepted:    181.1K
 * Total Submissions: 396.2K
 * Testcase Example:  '[["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]'
 *
 * 给你一个 m x n 的矩阵 board ，由若干字符 'X' 和 'O' ，找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X'
 * 填充。
 *
 *
 *
 *
 * 示例 1：
 *
 *
 * 输入：board =
 * [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
 * 输出：[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
 * 解释：被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O'
 * 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
 *
 *
 * 示例 2：
 *
 *
 * 输入：board = [["X"]]
 * 输出：[["X"]]
 *
 *
 *
 *
 * 提示：
 *
 *
 * m == board.length
 * n == board[i].length
 * 1
 * board[i][j] 为 'X' 或 'O'
 *
 *
 *
 *
 */

// @lc code=start
#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
    vector<int> direction{-1, 0, 1, 0, -1};
    void solve(vector<vector<char>> &board) {
        int m = board.size();
        if (m == 0) {
            return;
        }
        int n = board[0].size();
        if (m <= 2 || n <= 2) {
            return;
        }
        for (int i = 0; i < m; ++i) {
            dfs(board, i, 0, m, n);
            dfs(board, i, n - 1, m, n);
        }
        for (int i = 0; i < n; ++i) {
            dfs(board, 0, i, m, n);
            dfs(board, m - 1, i, m, n);
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == 'A') {
                    board[i][j] = 'O';
                }
                else if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
            }
        }
        return;
    }

    // helper
    void dfs(vector<vector<char>> &board, int r, int c, int m, int n) {
        if (r < 0 || r >= m || c < 0 || c >= n || board[r][c] != 'O') {
            return;
        }
        board[r][c] = 'A';
        int x, y;
        for (int i = 0; i < 4; ++i) {
            x = r + direction[i], y = c + direction[i + 1];
            dfs(board, x, y, m, n);
        }
    };
};
// @lc code=end
